Chapter 11, Problem 1RE

In Problems $1-4$, list the five terms of each sequence.

$\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

To determine

The first five terms of the sequence, whose $n^{th}$ term is $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

Answer to Problem 1RE

Solution:

The first five terms of the sequence are $-\frac{4}{3},\frac{5}{4},-\frac{6}{5},\frac{7}{6},-\frac{8}{7}$ respectively.

Explanation of Solution

Given information:

The sequence has $n^{th}$ term $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

Explanation:

To find the first term of the sequence, Plug $n=1$ in $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

$\Rightarrow a_1={\left(-1\right)}^1\left(\frac{1+3}{1+2}\right)$

$\Rightarrow a_1=-1\cdot \left(\frac{4}{3}\right)$

$\Rightarrow a_1=-\frac{4}{3}$

To find the second term, plug $n=2$ in $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

$\Rightarrow a_2={\left(-1\right)}^2\left(\frac{2+3}{2+2}\right)$

$\Rightarrow a_2=\frac{5}{4}$

To find the third term, plug $n=3$ in $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

$\Rightarrow a_3={\left(-1\right)}^3\left(\frac{3+3}{3+2}\right)$

$\Rightarrow a_3=\left(-1\right)\left(\frac{6}{5}\right)$

$\Rightarrow a_3=-\frac{6}{5}$

To find the fourth term, plug $n=4$ in $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

$\Rightarrow a_4={\left(-1\right)}^4\left(\frac{4+3}{4+2}\right)$

$\Rightarrow a_4=\frac{7}{6}$

To find the fifth term, plug $n=5$ in $\left\{a_n\right\}=\left\{{\left(-1\right)}^n\left(\frac{n+3}{n+2}\right)\right\}$

$\Rightarrow a_5={\left(-1\right)}^5\left(\frac{5+3}{5+2}\right)$

$\Rightarrow a_5=\left(-1\right)\left(\frac{8}{7}\right)$

$\Rightarrow a_5=-\frac{8}{7}$

Hence, the first five terms are $-\frac{4}{3},\frac{5}{4},-\frac{6}{5},\frac{7}{6}$ and $-\frac{8}{7}$